设A为三阶矩阵,Aj是A的第j列(j=1,2,3)矩阵B=(A3,3A2-A3,2A1+5A2),若|A|=-2.B等于多少

Question

像这一类的题都怎么做啊 有没有什么性质定理什么的...

Answer 1

| A3,3A2-A3,2A1+5A2 |
c2+c1

= | A3,3A2,2A1+5A2 |
= 3 | A3,A2,2A1+5A2 |
c3-5c2
= 3 | A3,A2,2A1 |
= 6 | A3,A2,A1 |

c1<->c3
= -6 | A1,A2,A3 |
= -6 |A|
= -6*(-2)
= 12.

高级解法就是转化为矩阵乘积的形式求行列式
B = (A3,3A2-A3,2A1+5A2) = (A1,A2,A3)K = AK
其中 K=
0 0 2
0 3 5
1 -1 0
|B| = |A||K| = -2 * (-6) = 12.